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Third Law of Thermodynamics

For the react...

Question

For the reaction $C H_{3} C O O H (l) + 2 O_{2} (g) ⇌ 2 C O_{2} (g) + 2 H_{2} (g) + 2 H_{2} O (l)$
at $25^{\circ} C$ and 1 atm pressure, $Δ H = - 874 K J$ Then the change in internal energy ( $Δ$ E) is

A

-874 KJ

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B

-871 .53 KJ

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C

-876.47 KJ

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D

+874 KJ

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Solution

The correct option is D -874 KJ
In this above reaction $Δ n$ = 0
So $Δ E = Δ H$
Hence A is correct option

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Similar questions

H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ 2 H_{2} O (l); Δ H = - 86 k J

2 H_{2} (g) + O_{2} (g) ⟶ 2 H_{2} O (l) . . . . . . . . . k J (\pm ?)

H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ 2 H_{2} O (l); Δ H = - 286 k J

2 H_{2} (g) + O_{2} (g) ⟶ 2 H_{2} O (l) . . . . . . . . . k J (\pm ?)

Q. For the reaction,

C_{2} H_{4} (g) + 3 O_{2} \to 2 C O_{2} (g) + 2 H_{2} O (l)

Δ U = - 1415

Δ H

27^{o}

Q. Given that :

2 C (s) + 2 O_{2} (g) \to 2 C O_{2} (g)

Δ H = - 787

H_{2} (g) + 1_{/ 2} O_{2} (g) \to H_{2} O (l)

Δ

= - 286

C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) \to 2 C O_{2} (g) + H_{2} O (l)

Δ

= - 1310

Heat of formation of acetylene is:

C (s) + O_{2} (g) \to C O_{2} (g)

Δ H = - 394

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C_{2} H_{5} O H (l) + 3 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (l)

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kJ/mole. Using the data, the heat of formation of ethanol is:

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